How to prove divergence of the integral $\int_{0}^{\infty}\frac{\sin(x)}{x^{2}}dx$

Question

I want to show that the following integral diverges:

$$\int_{0}^{\infty}\frac{\sin(x)}{x^{2}}dx$$

I used the substitution $ t = \frac{1}{x} $ to transform this integral into

$$\int_{0}^{\infty}\sin \frac{1}{t}dt$$

It seems intuitive that for large $t$, $\sin \frac{1}{t} = O(\frac{1}{t})$ and so the last integral diverges and so does the original. Is this correct?

Answer 1

After noting that $$\lim_{x \to 0^+} \frac{\sin x}{x} = 1$$

we have that $\sin x / x > 1/2$ for all $x$ sufficiently small. Hence the integrand can be bounded below via

$$\frac{\sin x}{x^2} > \frac 1 {2x}$$

for small enough $x$. Now compare.

Answer 2

Hint: Just note that as $x\sim 0$ the by Taylor series we have

$$ \frac{\sin x}{ x^2} \sim \frac{1}{x}. $$